Calculus
posted by Anonymous on .
Answer the following questions for the function
f(x) = sin^2(x/3)
defined on the interval [ 9.424778, 2.356194].
Rememer that you can enter pi for \pi as part of your answer.
a.) f(x) is concave down on the interval .
b.) A global minimum for this function occurs at .
c.) A local maximum for this function which is not a global
maximum occurs at .
d.) The function is increasing on the region .

We went through this already:
Posted by Anonymous on Tuesday, November 1, 2011 at 1:53pm.
f(x) = sin^2(x/2)
defined on the interval [ 5.683185, 1.270796].
Take a look at that one for the method. 
i cant see how you figure that out

I cant find the interval in which it concaves down, and that's the only question i need.

f = sin^2(x/3)
f' = 2/3 sin(x/3)cos(x/3) = 1/3 sin(2x/3)
f'' = 2/9 cos(2x/3)
f is concave down when f'' < 0
so, where is cos(2x/3) < 0?
When pi/2 < 2x/3 < 3pi/2
or 3pi/4 < x < 9pi/4
Your interval appears to be about (3pi/2,pi/2), so you may have to adjust your answer a bit to fit the interval.